Problem: Evaluate $~~\int x\sqrt{x+2}\,dx\,$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac23(x+2)^{3/2}(x-1)+C$ (Choice B) B $x\cdot\dfrac23(x+2)^{3/2}-\dfrac4{15}(x+2)^{5/2}+C$ (Choice C) C $x\cdot\dfrac23(x+2)^{3/2}-\dfrac2{3}(x+2)^{5/2}+C$ (Choice D) D $x\cdot\dfrac23(x+2)^{3/2}+\dfrac4{15}(x+2)^{5/2}+C$
We will solve this by integrating by parts. We know that $ \int u(x)v\,^\prime(x)dx = u(x)v(x)-\int u\,^\prime(x)v(x)dx\,$. We can rewrite this as $ \int u\ dv = uv-\int v\ du\,$. In this problem we will let $~u = x~$ and $~dv=\sqrt{x+2}\, dx\,$. Then $~du = dx~$ and $~v = \dfrac23(x+2)^{3/2}\,$. Integration by parts gives $ \int x\sqrt{x+2}\,dx = x\cdot\dfrac23(x+2)^{3/2}-\int\dfrac23(x+2)^{3/2}\,dx$ $ =x\cdot\dfrac23(x+2)^{3/2}-\dfrac23\cdot\dfrac25(x+2)^{5/2}+C$ $ =x\cdot\dfrac23(x+2)^{3/2}-\dfrac4{15}(x+2)^{5/2}+C\,$.